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Thread: algebra question: In printing an article of 48,000 words....

  1. #1
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    algebra question: In printing an article of 48,000 words....

    In printing an article of 48,000 words, a printer decides to use two sizes of type. Using the larger type, a printed page contains 1,800 words. Using smaller type, a page contains 2,400 words. The article is allotted 21 full pages in a magazine. How
    many pages must be in smaller type?

  2. #2
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    Re: algebra question

    Quote Originally Posted by atwana_b
    in printing an article of 48,000 words, a printer decides to use two sizes of type. using the larger type, a printed page contains 1,800 words. using smaller type, a page contains 2,400 words. the article is allotted 21 full pages in a magazine. how
    many pages must be in smaller type.
    Start by NAMING things.

    I might do this:

    let x = # of pages in large type
    let y = # pages in small type

    We knkow that there are to be 21 pages in all, so
    x + y = 21

    We also know that
    words on small type pages + words on large type pages = 48000
    since the article contains 48000 words.

    Using large type, there are 1800 words per page, so x pages of large type will hold 1800x words.

    Using small type, there are 2400 words per page, so y pages of small type will hold 2400 words.

    Together, the two type sizes must hold 48000 words. So,
    1800x + 2400y = 48000

    Ok...now you have a system of two equations in two variables:

    x + y = 21
    1800x + 2400y = 48000

    Solve this system....

    If you have trouble solving the sysem, please repost showing your work.

  3. #3
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    algebra question

    Ok, I am horrible at word problems --

    1800x + 2400y = 48000

    so would I do this:

    1800x = 48000 (divide both sides by 1800) and then x=26.66

    2400y = 48000 (divide both sides by 2400) and then y=20

    If I did start off right, I am not sure what to do next........

  4. #4
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    Re: algebra question

    Quote Originally Posted by atwana_b
    Ok, I am horrible at word problems --
    1800x + 2400y = 48000
    so would I do this:
    1800x = 48000 (divide both sides by 1800) and then x=26.66
    2400y = 48000 (divide both sides by 2400) and then y=20
    If I did start off right, I am not sure what to do next........
    If that's what you did after being given this by Mrs. Pi:

    Quote Originally Posted by Mrspi
    x + y = 21
    1800x + 2400y = 48000
    Solve this system....
    ...then you need to see your teacher; we cannot teach here.
    I'm just an imagination of your figment !

  5. #5
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    algebra question

    I am sorry that you feel that way -- but thanks to Tchrwill & Eliz, I was taught here.

    Is there anyone who would be willing to help me understand how to solve this equation? It would really mean a lot.

  6. #6
    Elite Member stapel's Avatar
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    Quote Originally Posted by atwana_b
    Is there anyone who would be willing to help me understand how to solve this equation?
    As was mentioned earlier, this is not one equation; it is a system of (two) equations. Since we cannot teach courses here, please use the lessons already available online.

    Thank you.

    Eliz.

  7. #7
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    algebra question

    Once again, you have helped me! Thank you so much

    1800x + 2400y = 48000

    1800 (-y+21) + 2400y = 48000
    -1800y + 37800 + 2400y = 48000
    600y + 37800= 48000
    600y = 10200
    y=17

    x + 17 = 21
    x = 4

    Have a happy holiday!

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